Applying Differentiation Rules To Logarithmic Functions
In this wiki, we will learn about differentiating logarithmic functions which are given by \(y=\log_{a} x\) , in particular the natural logarithmic function \(y=\ln x\) using the differentiation rules . We can easily prove that these logarithmic functions are easily differentiable by looking at there graphs:
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Image Source: ltcconline.net
General Result
Prove that
\[\dfrac{d}{dx} \log_{a} x = \dfrac{1}{x\ln a}.\]
Proof: Let \(y=\log_{a} x,\) then by basic logarithmic properties we have
\[a^y=x.\]
Differentiating the above equation implicitly with respect to \(x\) and using \(\dfrac{d}{dx}a^x=a^x\ln a,\) we get
\[\begin{align} a^y\ln a\dfrac{dy}{dx}&=1\\ \Rightarrow \dfrac{dy}{dx}&=\dfrac{1}{a^y\ln a}\\&=\dfrac{1}{x\ln a}.\ _\square \end{align}\]
Special Case of the General Result
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If we put \(a=e\) in the general result where \(e\) is Euler's number, we have \(\ln a=\ln e=1\) and get an amazing result:
\[\dfrac{d}{dx} \ln x=\dfrac{1}{x}.\]
But we have a more worth-remembering result than the above one:
Prove that
\[\dfrac{d}{dx} \ln |x|=\dfrac{1}{x}.\]
Proof: Using the property of absolute value , we know the following about the function \(f(x)=\ln|x|:\)
\[ f(x)=\begin{cases} \ln x \ & \text{if} \ & x>0 \\ \ln(-x) \ & \text{if} \ & x<0. \end{cases}\]
Now, it follows that
\[ f'(x)=\begin{cases} \dfrac{1}{x} \ & \text{if} \ & x>0 \\ \dfrac{1}{-x} (-1)=\dfrac{1}{x} \ & \text{if} \ & x<0. \end{cases}\]
Therefore,
\[f'(x)=\dfrac{1}{x} \implies \dfrac{d}{dx} \ln|x|=\dfrac{1}{x} \ \forall \ x \neq 0.\ _\square\]
Example Problems
Prove the power rule using logarithmic differentiation.
Proof:
First we state the power rule: for any real number \(n\),
\[ \frac{d}{dx} x^n = n x ^{n-1 }. \]
So first, we let \(y=x^n,\) then by logarithmic differentiation for \(x\neq 0\) we have
\[\ln|y|=\ln|x^n|=n\ln |x|.\]
Using \(\frac{d}{dx} \ln|x|=\frac{1}{x}\) (as proved above), we have
\[\begin{align} \dfrac{y'}{y}&=\dfrac{n}{x}\\ \Rightarrow y'&=\dfrac{ny}{x}\\&=\dfrac{nx^{n}}{x}\\&=nx^{n-1}, \end{align}\]
which was to be proved. \(_\square\)
Differentiate \(y=x^{\sqrt{x}}.\)
We can write \(x=e^{\ln x}\) using logarithmic rules, and thus we have
\[x^{\sqrt{x}}=\left(e^{\ln x}\right)^{\sqrt{x}}.\]
Differentiating both the sides, we have
\[\dfrac{d}{dx} x^{\sqrt{x}}=\dfrac{d}{dx} e^{\sqrt{x}\ln x}.\]
Using chain rule, we have
\[\dfrac{d}{dx} e^{\sqrt{x}\ln x}=e^{\sqrt{x}\ln x}\dfrac{d}{dx} \sqrt{x}\ln x = x^{\sqrt{x}}\dfrac{d}{dx} \sqrt{x}\ln x. \]
Using product rule, we have
\[\dfrac{d}{dx} \sqrt{x}\ln x = \sqrt{x} \cdot \dfrac{1}{x} + \ln x \cdot \dfrac{1}{2\sqrt{x}}=\dfrac{2+\ln x}{2\sqrt{x}}. \]
Therefore, we have
\[\dfrac{d}{dx} x^{\sqrt{x}}= x^{\sqrt{x}}\left(\dfrac{2+\ln x}{2\sqrt{x}}\right).\ _\square \]
Differentiate \(y=2e^x+5x^2\ln x.\)
First, let us try to differentiate the second term \(5x^2\ln x.\)
Using product rule,
\[\dfrac{d}{dx} 5x^2\ln x = 10x\ln x+5x^2\cdot\dfrac{1}{x}=10x\ln x+5x.\]
Thus, we have
\[\dfrac{dy}{dx}=2e^x+10x\ln x+5x,\]
which is what we needed to find. \(_\square\)
Differentiate \(y=\dfrac{2e^x}{3e^x+1}.\)
This is where we need to directly use the quotient rule.
Using quotient rule, we have
\[\begin{align} \dfrac{dy}{dx} &=\dfrac{2e^x(3e^x+1)-(3e^x)(2e^x)}{(3e^x+1)^2}\\ &= \dfrac{6e^{2x}+2e^x-6e^{2x}}{(3e^x+1)^2}\\ \Rightarrow \dfrac{dy}{dx}&=\dfrac{2e^x}{(3e^x+1)^2}.\ _\square \end{align}\]
Differentiate \(y=x^x\) for \(x>0.\)
We cannot directly approach this using differentiation rules. We need to bring suitable form for the function to be differentiated:
\[y=x^x\implies \ln y=\ln x^x \implies \ln y= x\ln x.\]
We now differentiate both sides with respect to \(x,\) using the chain rule on the left side and the product rule on the right:
\[\begin{align} \dfrac{dy}{dx} \cdot \dfrac{1}{y} &= \ln x + x \cdot\dfrac{1}{x}\\ &=\ln x + 1\\ \Rightarrow \dfrac{dy}{dx} &= y(\ln x + 1) \\ &= x^x(\ln x+1).\ _\square \end{align}\]
Community Problems